3.303 \(\int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=548 \[ -\frac {2 f^2 \left (a^2-b^2\right ) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 f^2 \left (a^2-b^2\right ) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {f (e+f x) \sin (c+d x) \cos (c+d x)}{2 b d^2}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}+\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d} \]

[Out]

1/2*e*f*x/b/d+1/4*f^2*x^2/b/d+1/3*I*(a^2-b^2)*(f*x+e)^3/b^3/f+2*a*f*(f*x+e)*cos(d*x+c)/b^2/d^2-(a^2-b^2)*(f*x+
e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d-(a^2-b^2)*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^
2)^(1/2)))/b^3/d+2*I*(a^2-b^2)*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^2+2*I*(a^2-b^
2)*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^2-2*(a^2-b^2)*f^2*polylog(3,I*b*exp(I*(d*
x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^3-2*(a^2-b^2)*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^3-2
*a*f^2*sin(d*x+c)/b^2/d^3+a*(f*x+e)^2*sin(d*x+c)/b^2/d-1/2*f*(f*x+e)*cos(d*x+c)*sin(d*x+c)/b/d^2+1/4*f^2*sin(d
*x+c)^2/b/d^3-1/2*(f*x+e)^2*sin(d*x+c)^2/b/d

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Rubi [A]  time = 0.73, antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4525, 3296, 2637, 4404, 3310, 4519, 2190, 2531, 2282, 6589} \[ \frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^2}-\frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^3}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \sin (c+d x) \cos (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}+\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^2*Cos[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

(e*f*x)/(2*b*d) + (f^2*x^2)/(4*b*d) + ((I/3)*(a^2 - b^2)*(e + f*x)^3)/(b^3*f) + (2*a*f*(e + f*x)*Cos[c + d*x])
/(b^2*d^2) - ((a^2 - b^2)*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 -
b^2)*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d) + ((2*I)*(a^2 - b^2)*f*(e + f*x
)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^2) + ((2*I)*(a^2 - b^2)*f*(e + f*x)*PolyLog[
2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) - (2*(a^2 - b^2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)
))/(a - Sqrt[a^2 - b^2])])/(b^3*d^3) - (2*(a^2 - b^2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2
])])/(b^3*d^3) - (2*a*f^2*Sin[c + d*x])/(b^2*d^3) + (a*(e + f*x)^2*Sin[c + d*x])/(b^2*d) - (f*(e + f*x)*Cos[c
+ d*x]*Sin[c + d*x])/(2*b*d^2) + (f^2*Sin[c + d*x]^2)/(4*b*d^3) - ((e + f*x)^2*Sin[c + d*x]^2)/(2*b*d)

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4525

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol
] :> Dist[a/b^2, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] + (-Dist[1/b, Int[(e + f*x)^m*Cos[c + d*x]^(n -
2)*Sin[c + d*x], x], x] - Dist[(a^2 - b^2)/b^2, Int[((e + f*x)^m*Cos[c + d*x]^(n - 2))/(a + b*Sin[c + d*x]), x
], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 1] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \int (e+f x)^2 \cos (c+d x) \, dx}{b^2}-\frac {\int (e+f x)^2 \cos (c+d x) \sin (c+d x) \, dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}-\frac {\left (a^2-b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac {\left (a^2-b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac {(2 a f) \int (e+f x) \sin (c+d x) \, dx}{b^2 d}+\frac {f \int (e+f x) \sin ^2(c+d x) \, dx}{b d}\\ &=\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}+\frac {f \int (e+f x) \, dx}{2 b d}+\frac {\left (2 \left (a^2-b^2\right ) f\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d}+\frac {\left (2 \left (a^2-b^2\right ) f\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d}-\frac {\left (2 a f^2\right ) \int \cos (c+d x) \, dx}{b^2 d^2}\\ &=\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}-\frac {\left (2 i \left (a^2-b^2\right ) f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^2}-\frac {\left (2 i \left (a^2-b^2\right ) f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^2}\\ &=\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}-\frac {\left (2 \left (a^2-b^2\right ) f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 d^3}-\frac {\left (2 \left (a^2-b^2\right ) f^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b^3 d^3}\\ &=\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}\\ \end {align*}

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Mathematica [B]  time = 5.97, size = 2397, normalized size = 4.37 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^2*Cos[c + d*x]^3)/(a + b*Sin[c + d*x]),x]

[Out]

((48*I)*a^2*d^3*e^2*E^((2*I)*c)*x - (48*I)*b^2*d^3*e^2*E^((2*I)*c)*x + (48*I)*a^2*d^3*e*E^((2*I)*c)*f*x^2 - (4
8*I)*b^2*d^3*e*E^((2*I)*c)*f*x^2 + (16*I)*a^2*d^3*E^((2*I)*c)*f^2*x^3 - (16*I)*b^2*d^3*E^((2*I)*c)*f^2*x^3 - (
48*I)*a^2*d^2*e^2*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + (48*I)*b^2*d^2*e^
2*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + (24*I)*a*b*d^2*e^2*E^(I*c)*Cos[d*
x] - (24*I)*a*b*d^2*e^2*E^((3*I)*c)*Cos[d*x] + 48*a*b*d*e*E^(I*c)*f*Cos[d*x] + 48*a*b*d*e*E^((3*I)*c)*f*Cos[d*
x] - (48*I)*a*b*E^(I*c)*f^2*Cos[d*x] + (48*I)*a*b*E^((3*I)*c)*f^2*Cos[d*x] + (48*I)*a*b*d^2*e*E^(I*c)*f*x*Cos[
d*x] - (48*I)*a*b*d^2*e*E^((3*I)*c)*f*x*Cos[d*x] + 48*a*b*d*E^(I*c)*f^2*x*Cos[d*x] + 48*a*b*d*E^((3*I)*c)*f^2*
x*Cos[d*x] + (24*I)*a*b*d^2*E^(I*c)*f^2*x^2*Cos[d*x] - (24*I)*a*b*d^2*E^((3*I)*c)*f^2*x^2*Cos[d*x] + 6*b^2*d^2
*e^2*Cos[2*d*x] + 6*b^2*d^2*e^2*E^((4*I)*c)*Cos[2*d*x] - (6*I)*b^2*d*e*f*Cos[2*d*x] + (6*I)*b^2*d*e*E^((4*I)*c
)*f*Cos[2*d*x] - 3*b^2*f^2*Cos[2*d*x] - 3*b^2*E^((4*I)*c)*f^2*Cos[2*d*x] + 12*b^2*d^2*e*f*x*Cos[2*d*x] + 12*b^
2*d^2*e*E^((4*I)*c)*f*x*Cos[2*d*x] - (6*I)*b^2*d*f^2*x*Cos[2*d*x] + (6*I)*b^2*d*E^((4*I)*c)*f^2*x*Cos[2*d*x] +
6*b^2*d^2*f^2*x^2*Cos[2*d*x] + 6*b^2*d^2*E^((4*I)*c)*f^2*x^2*Cos[2*d*x] - 24*a^2*d^2*e^2*E^((2*I)*c)*Log[4*a^
2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + 24*b^2*d^2*e^2*E^((2*I)*c)*Log[4*a^2*E^((2*I)*(c +
d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - 96*a^2*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E
^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 96*b^2*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^
(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 48*a^2*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E
^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 48*b^2*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*
E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 96*a^2*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E
^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 96*b^2*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^
(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 48*a^2*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E
^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 48*b^2*d^2*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*
E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (96*I)*(a^2 - b^2)*d*E^((2*I)*c)*f*(e + f*x)*PolyLog[2, (I*b*E^(I
*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (96*I)*(a^2 - b^2)*d*E^((2*I)*c)*f*(e + f*x)*
PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 96*a^2*E^((2*I)*c)*f^2*P
olyLog[3, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 96*b^2*E^((2*I)*c)*f^2*Pol
yLog[3, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 96*a^2*E^((2*I)*c)*f^2*PolyL
og[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 96*b^2*E^((2*I)*c)*f^2*PolyLo
g[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 24*a*b*d^2*e^2*E^(I*c)*Sin[d*x
] + 24*a*b*d^2*e^2*E^((3*I)*c)*Sin[d*x] - (48*I)*a*b*d*e*E^(I*c)*f*Sin[d*x] + (48*I)*a*b*d*e*E^((3*I)*c)*f*Sin
[d*x] - 48*a*b*E^(I*c)*f^2*Sin[d*x] - 48*a*b*E^((3*I)*c)*f^2*Sin[d*x] + 48*a*b*d^2*e*E^(I*c)*f*x*Sin[d*x] + 48
*a*b*d^2*e*E^((3*I)*c)*f*x*Sin[d*x] - (48*I)*a*b*d*E^(I*c)*f^2*x*Sin[d*x] + (48*I)*a*b*d*E^((3*I)*c)*f^2*x*Sin
[d*x] + 24*a*b*d^2*E^(I*c)*f^2*x^2*Sin[d*x] + 24*a*b*d^2*E^((3*I)*c)*f^2*x^2*Sin[d*x] - (6*I)*b^2*d^2*e^2*Sin[
2*d*x] + (6*I)*b^2*d^2*e^2*E^((4*I)*c)*Sin[2*d*x] - 6*b^2*d*e*f*Sin[2*d*x] - 6*b^2*d*e*E^((4*I)*c)*f*Sin[2*d*x
] + (3*I)*b^2*f^2*Sin[2*d*x] - (3*I)*b^2*E^((4*I)*c)*f^2*Sin[2*d*x] - (12*I)*b^2*d^2*e*f*x*Sin[2*d*x] + (12*I)
*b^2*d^2*e*E^((4*I)*c)*f*x*Sin[2*d*x] - 6*b^2*d*f^2*x*Sin[2*d*x] - 6*b^2*d*E^((4*I)*c)*f^2*x*Sin[2*d*x] - (6*I
)*b^2*d^2*f^2*x^2*Sin[2*d*x] + (6*I)*b^2*d^2*E^((4*I)*c)*f^2*x^2*Sin[2*d*x])/(48*b^3*d^3*E^((2*I)*c))

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fricas [C]  time = 0.65, size = 1793, normalized size = 3.27 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(b^2*d^2*f^2*x^2 + 2*b^2*d^2*e*f*x + 4*(a^2 - b^2)*f^2*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x +
c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 4*(a^2 - b^2)*f^2*polylog(3, 1/2*(2*I
*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 4*(a^2
- b^2)*f^2*polylog(3, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqr
t(-(a^2 - b^2)/b^2))/b) + 4*(a^2 - b^2)*f^2*polylog(3, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(
d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - (2*b^2*d^2*f^2*x^2 + 4*b^2*d^2*e*f*x + 2*b^2*d^2*e^2
- b^2*f^2)*cos(d*x + c)^2 - 8*(a*b*d*f^2*x + a*b*d*e*f)*cos(d*x + c) - (-4*I*(a^2 - b^2)*d*f^2*x - 4*I*(a^2 -
b^2)*d*e*f)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(
a^2 - b^2)/b^2) + 2*b)/b + 1) - (-4*I*(a^2 - b^2)*d*f^2*x - 4*I*(a^2 - b^2)*d*e*f)*dilog(-1/2*(2*I*a*cos(d*x +
c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - (4*I*(a^
2 - b^2)*d*f^2*x + 4*I*(a^2 - b^2)*d*e*f)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x +
c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1) - (4*I*(a^2 - b^2)*d*f^2*x + 4*I*(a^2 - b^2)*d*e*f
)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)
/b^2) + 2*b)/b + 1) + 2*((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^2)*log(2*b*cos(d*x +
c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f
+ (a^2 - b^2)*c^2*f^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*((
a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) +
2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^2)*l
og(-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*((a^2 - b^2)*d^2*f^2*x^2 +
2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(
d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 - b^2)*d^2*f^2*x^2
+ 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*si
n(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 - b^2)*d^2*f^2*x
^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a
*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) + 2*((a^2 - b^2)*d^2*f^
2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^2)*log(1/2*(-2*I*a*cos(d*x + c) +
2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(2*a*b*d^2*f^2*x
^2 + 4*a*b*d^2*e*f*x + 2*a*b*d^2*e^2 - 4*a*b*f^2 - (b^2*d*f^2*x + b^2*d*e*f)*cos(d*x + c))*sin(d*x + c))/(b^3*
d^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \cos \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*cos(d*x + c)^3/(b*sin(d*x + c) + a), x)

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maple [F]  time = 2.68, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

int((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(e + f*x)^2)/(a + b*sin(c + d*x)),x)

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*cos(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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